What is the First Derivative of Kinetic Energy?

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I had a question on the first derivative of Kinetic Energy...This a problem from the FE exam...

Now the solution is = mva...which turns out to be power...im just trying to understand how they came up with that...

so KE = 1/2 mv^2


im asuming they are using the general power rule so v^2 would be 2v dv/dt where dv/dt is acceleration...

the m is where I am confused...the derivative of m would just be 1...but why not dm/dt as well...obviously it wouldn't make sense because the mass wouldn't change with time...but i guess I am rusty on the power rule...
 
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Jason03 said:
I had a question on the first derivative of Kinetic Energy...This a problem from the FE exam...

Now the solution is = mva...which turns out to be power...im just trying to understand how they came up with that...

so KE = 1/2 mv^2


im asuming they are using the general power rule so v^2 would be 2v dv/dt where dv/dt is acceleration...

the m is where I am confused...the derivative of m would just be 1...but why not dm/dt as well...obviously it wouldn't make sense because the mass wouldn't change with time...but i guess I am rusty on the power rule...


m is a constant, v is the only thing changing with time, that's why you differentiate the v2 with respect to t
 
perfect!...that makes sense...I thought it was something to do with mass not changing...but i forgot i was taking the derivative with respect ot time...

thank you
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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